`int_(0)^(a) (dx)/(x+ sqrt""(a^(2) -x^(2)))`=

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `pi`
D. none

Answer

To solve the integral \( I = \int_{0}^{a} \frac{dx}{x + \sqrt{a^2 - x^2}} \), we will use a trigonometric substitution. Here are the steps:

### Step 1: Trigonometric Substitution
Let \( x = a \cos \theta \). Then, we differentiate to find \( dx \):
\[ dx = -a \sin \theta \, d\theta \]

### Step 2: Change the Limits of Integration
When \( x = 0 \):
\[ 0 = a \cos \theta \implies \cos \theta = 0 \implies \theta = \frac{\pi}{2} \]
When \( x = a \):
\[ a = a \cos \theta \implies \cos \theta = 1 \implies \theta = 0 \]
Thus, the limits change from \( x = 0 \) to \( x = a \) into \( \theta = \frac{\pi}{2} \) to \( \theta = 0 \).

### Step 3: Substitute into the Integral
Substituting \( x \) and \( dx \) into the integral:
\[ I = \int_{\frac{\pi}{2}}^{0} \frac{-a \sin \theta \, d\theta}{a \cos \theta + \sqrt{a^2 - (a \cos \theta)^2}} \]
Simplifying the square root:
\[ \sqrt{a^2 - a^2 \cos^2 \theta} = \sqrt{a^2(1 - \cos^2 \theta)} = a \sin \theta \]
Thus, the integral becomes:
\[ I = \int_{\frac{\pi}{2}}^{0} \frac{-a \sin \theta \, d\theta}{a \cos \theta + a \sin \theta} = \int_{\frac{\pi}{2}}^{0} \frac{-a \sin \theta \, d\theta}{a(\cos \theta + \sin \theta)} \]
This simplifies to:
\[ I = -\int_{\frac{\pi}{2}}^{0} \frac{\sin \theta \, d\theta}{\cos \theta + \sin \theta} \]

### Step 4: Reverse the Limits
Reversing the limits changes the sign:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta \, d\theta}{\cos \theta + \sin \theta} \]

### Step 5: Add Two Integrals
Let \( J = \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta \, d\theta}{\cos \theta + \sin \theta} \).
Now, we can add \( I \) and \( J \):
\[ I + J = \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta + \cos \theta}{\cos \theta + \sin \theta} \, d\theta = \int_{0}^{\frac{\pi}{2}} d\theta = \frac{\pi}{2} \]

### Step 6: Solve for \( I \)
Since \( I = J \):
\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4} \]

Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{4}} \]